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University of Oklahoma × + O saplingleaming.com/ibiscms/mod/ibi5Nie w.php?id-3871 622 Sapling Learning macmillan learning Jump to... Andrea Brisby Question 2 of 15 Available l Due Date: Points Pos Grade Cat Descriptio Map dot General Chemistry 4th Edition University Science Books presented by Sapling Learning A solution of household bleach contains 5.25% sodium hypochlorite, NaOCl, by mass. Assuming that the density of bleach is the same as water, calculate the volume of household bleach that should be diluted with water to make 500.0 mL of a pH 9.80 solution. Use the Ka of hypochlorous acid found in the chempendix 5 Solutions You can You can You have There is 8 eTextbo 10 Help Wi 12 3 O Web He 2011-2017 Sapling Learning, Inc. about uscareersprivacy policy terms of use contact us help 11:47 AM Type here to search ^4x 11/3/2017 ea

Explanation / Answer

C = %.25% NaOCl

d = 1 g/mL

find volume of this solution to make a dilution V = 500 mL, pH = 9.80

expect hydrolysis

ClO-(aq) + H2O(l) <-> HClO+ OH-(aq)

Let HA --> HClO and A- = ClO- for simplicity

since A- is formed

the next equilibrium is formed, the conjugate acid and water

A- + H2O <-> HA + OH-

The equilibrium s best described by Kb, the base constant

Kb by definition since it is an base:

Kb = [HA ][OH-]/[A-]

Ka can be calculated as follows:

Kb = Kw/Ka = (10^-14)/(4*10^-8) = 2.5*10^-7

get ICE table:

Initially

[OH-] = 0

[HA] = 0

[A-] = M

the Change

[OH-] = + x

[HA] = + x

[A-] = - x

in Equilibrium

[OH-] = 0 + x

[HA] = 0 + x

[A-] = M - x

get X from

pOH = 14-pH = 14-9.80 = 4.2

[OH-] =10^-4.2 = 0.0000630

[HA] = 0.0000630

substitute in Kb expression

Kb = [HA ][OH-]/[A-]

2.5*10^-7 = 0.0000630*0.0000630/(M-0.0000630)

solve for M

M = (0.0000630*0.0000630) / (2.5*10^-7)+0.0000630

M = 0.015939 mol of OCl- / liter required in V = 500 mL

mol of OCl- = MV = 0.015939*0.5 = 0.0079695 mol of OCl-

mol of NaOCl reuqired = 0.0079695mol

mass of NaCl = mol*MW = 0.0079695*74.44 = 0.59324 g

from original solution

C = mass/V

V = mass/C

V = (0.59324)/(0.0525) = 11.299 mL = 11.3 mL required

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